Problem 1

We say that a triangle

ABC

is great if the following holds: for any point

D

on side

BC

, if

P

and

Q

are the feet of the perpendiculars from

D

to the lines

AB

and

AC

respectively, then the reflection of

D

in the line

PQ

lies on the circumcircle of triangle

ABC

. Prove that triangle

ABC

is great if and only if

\angle A=90^\circ

and

AB=AC

Step 4 of 5: Midpoint of

BC

forces

AB=AC

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D=\text{midpoint of }BC,\ \angle BAC=90^\circ \implies PQ\parallel BC,\ DD'\perp BC \implies D'=A \implies AB=AC

Detailed analysis

Now let $D$ be the midpoint of $BC$ instead. Because $\angle BAC=90^\circ$ , triangle $DQP$ (with $P$ , $Q$ the feet of the perpendiculars from this new $D$ ) is the medial triangle of $ABC$ , so $PQ$ is parallel to $BC$ and therefore $DD'$ is perpendicular to $BC$ . The distance from $D'$ to $BC$ equals both the circumradius of $ABC$ and the distance from $A$ to $BC$ ; this forces $D'=A$ , which is only possible when $ABC$ is isosceles, i.e. $AB=AC$ . Together with $\angle BAC=90^\circ$ , this proves the forward direction.